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$\int \frac{\log_{e}x}{x\sqrt{1+\log_{e}x}}.dx=$

Option 1)
$(1+\log_{e}x)^{\frac{3}{2}}+c$

Option 2)
$\frac{2}{3}(1+\log_{e}x)(\log_{e}x-2)+c$

Option 3)
$\frac{2}{3}(1+l_nx)^{\frac{1}{2}}(l_nx-5)+c$

Option 4)
$\frac{2}{3}(1+l_nx)^{\frac{1}{2}}(l_nx-2)+c$

Answers (1)

best_answer

As learnt

Type of integration by substitution -

$\int \frac{f'(x)dx}{\sqrt{f(x)}}=2\sqrt{f(x)}+c$

- wherein

Let $f(x)=t$

$\therefore f{}'(x)dx=dt$

$\int \frac{\log_{e}x}{\sqrt[x]{1+ \log_{e}x}} \: dx$ , Let $\\ 1+\log_{e}x=t \\ \log_{e}x=t-1 \: \: , \: \: \frac{1}{x}dx=dt$

$\int \frac{-1+t dt}{\sqrt{t}} \: \: = \: \int \frac{t}{\sqrt{t}}dt-\int \frac{1}{\sqrt{t}}dt$

$\\ = \int \sqrt{t}dt-\int \frac{dt}{\sqrt{t}} \\ \\ \Rightarrow \frac{2}{3}\left ( 1+\log_{e}x \right )^{\frac{3}{2}}-2(1+\log_{e}x)^{\frac{1}{2}}$

$=2(\log_{e}x+1)^{\frac{1}{2}}\left [ \frac{(1+\log_{e}x)}{3}-1 \right ]$

$=\frac{2}{3}\left ( 1+\log_{e}x \right )^{\frac{1}{2}}(\log_{e}x-2)+c$

Option 1)

$(1+\log_{e}x)^{\frac{3}{2}}+c$

This option is incorrect

Option 2)

$\frac{2}{3}(1+\log_{e}x)(\log_{e}x-2)+c$

This option is incorrect

Option 3)

$\frac{2}{3}(1+l_nx)^{\frac{1}{2}}(l_nx-5)+c$

This option is incorrect

Option 4)

$\frac{2}{3}(1+l_nx)^{\frac{1}{2}}(l_nx-2)+c$

This option is correct

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